Summer Induction Work

Transcription

1 Further Maths Summer Induction Work Deadline: Friday 7th September The Hazeley Academy

2 Further Mathematics OCR (MEI): Further Core Pure, Mechanics, Statistics Objectives: To reinforce understanding of key GCSE and other mathematical concepts needed for success in A-level Further Maths. Tasks The topics in this booklet are extension of the ones covered in the A-level Maths booklet, Keeping your Maths brain, so you might need it when completing these tasks. There are 3 main topics in the booklet: Quadratics, Trigonometry and Mathematical Proof & Notation. Each one starts with useful notes and examples. You should read through the notes and work through the examples to ensure that you fully understand the concepts before moving on to the exercises. Each exercise should take about an hour although you might need to spend more or less time depending on how you find the individual topics. You might not be able to do the questions immediately but don t give up, think carefully about the information given and refer to the examples to help you. You must attempt all the questions, showing full working out. Remember to check your answers as you go and highlight any areas that you think you need more help with when you come back in September. The deadline for completing and handing in this homework booklet is 7 th September. The homework will be marked and graded according the following scheme. Grade A B C D E U Grade threshold 80% 70% 60% 50% 40% Below 40% Resources/Research Use the notes provided in the A-level maths booklet along with resources such as and to support your learning. Wider Reading If you are interested in some wider reading, the following books may be of interest: Fermat s Last Theorem Simon Singh The Music of the Primes: Why an Unsolved Problem in Mathematics Matters: Marcus Du Sautoy Seventeen Equations that Changed the World Ian Stewart Who Tells the Truth? Collection of Logical Puzzles to Make You Think Adam Case The Great Mathematical Problems: Marvels and Mysteries of Mathematics Ian Stewart Submission Date 7 th September 018

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4 AS Mathematics Quadratic functions Section : The quadratic formula Notes and Examples These notes contain subsections on Solving quadratic equations using the formula Problem solving Solving quadratic equations using the formula If a quadratic equation is written in the completed square form, it is easy to solve. Example 1 (i) Write x 4x 5 in the completed square form. (ii) Hence solve the equation x 4x 5 0 Solution (i) x 4x 5 ( x ) 4 5 ( x ) 9 (ii) ( x ) 9 x 3 x 3 x 5 or 1 However, unless you already have the equation in the completed square form, as in the example above, it is easier to use the quadratic formula, which is just a generalisation of the completing the square method. The quadratic formula for the solutions of the equation ax² bx c 0 is b b² 4ac x. a The expression b² 4ac is called the discriminant. This is very important as it tells you something about the nature of the solutions. In each case the solution(s) correspond to the points where the graph meets the x-axis. If the discriminant is positive, then there are two real solutions. (If the discriminant is a positive square number, then the two real solutions are rational and it is possible to solve the equation by factorisation; otherwise the solutions are irrational and you must use the quadratic formula.) 1 of 4 1/07/16 MEI

5 AS Maths Quadratics Notes and Examples y = x² + x 3 Discriminant = 16 Two rational solutions y = x² + x 3 Discriminant = 13 Two real, irrational solutions If the discriminant is zero, then the quadratic is a perfect square and there is one real solution, which can be found by factorisation. y = x² + x + 1 Discriminant = 0 One real solution If the discriminant is negative, then there are no real solutions. y = x² + x + Discriminant = -4 No real solutions As the graph does not meet the x-axis, there cannot be any real solutions. When you need to solve a quadratic equation, it is useful to quickly work out the discriminant before you start, so that you know whether there are real solutions, and whether the equation can be solved by factorisation. Your calculator may be able to solve quadratic equations, and some calculators will give the answers in exact form (using surds). However, you may be required to show working in some questions, so you must know the quadratic formula and be confident in using it. of 4 1/07/16 MEI

6 AS Maths Quadratics Notes and Examples Example For each of the following quadratic equations, find the discriminant and solve the equation, where possible, by a suitable method (i) x² 4x1 0 (ii) 6x ² 11x 10 0 (iii) 3x ² x 4 0 (iv) 4x ² 1x 9 0 Solution (i) a =, b = -4, c = 1 Discriminant = ( 4)² Since the discriminant is positive, there are two real solutions. As it is not a square number, the equation must be solved using the quadratic formula. b b² 4ac x a (ii) a = 6, b = 11, c = -10 Discriminant = 11² Since the discriminant is positive, there are two real solutions. As it is a square number (19²), the equation can be solved by factorisation. 6x² 11x 10 0 (3x )(x 5) 0 x 3 or x 5 (iii) a = 3, b = -, c = 4 Discriminant = ( )² Since the discriminant is negative, there are no real solutions. (iv) a = 4, b = 1, c = 9 Discriminant = 1² Since the discriminant is zero, there is one solution and the equation can be solved by factorisation into a perfect square. 4x² 1x 9 0 (x 3)² 0 x 3 Try the Quadratic formula walkthrough, and the Quadratic formula skill pack. 3 of 4 1/07/16 MEI

7 AS Maths Quadratics Notes and Examples Problem solving Some problems, when translated into algebra, involve quadratic equations. Example 3 A rectangular box has width cm greater than its length, and height 3 cm less than its length. The total surface area of the box is 548 cm². What are the dimensions of the box? Solution Let the length of the box be x cm. The width of the box is x + cm, and the height is x 3 cm. The surface area of the box is given by x( x ) x( x 3) ( x )( x 3). x( x ) x( x 3) ( x )( x 3) 548 x( x ) x( x 3) ( x )( x 3) 74 x x x x x x x x 80 0 (3x 8)( x10) 0 x 10 Divide through by The discriminant is 3364, which is 58², so this must factorise 3x + 8 = 0 gives a negative value of x, which does not make sense in this context. So the solution must be x 10 = 0. The length of the box is 10 cm, the width is 1 cm and the height is 7 cm. Notice that in Example 3, you could discard one of the possible solutions as a negative solution did not make sense in the context. This is not always the case. In some situations, a negative solution can have a practical meaning. For example if the height of a stone thrown from the edge of a cliff is negative, this simply means that the stone is below the level of the cliff at that point. However, if the stone was thrown from level ground, then a negative height does not make sense. Some problems leading to quadratic equations do have two possible solutions. Always consider whether your solution(s) make sense in the context. 4 of 4 1/07/16 MEI

8 MEI AS Mathematics Quadratic functions Topic assessment 1. Solve each of the following quadratic equations, if possible, giving answers in exact form. (i) x x 3 0 (ii) 3x x 4 0 (iii) x 5x1 0 [5]. (i) Write the quadratic expression x 4x 5 in the form A( x B) C. [] (ii) Find the discriminant of the quadratic equation x 4x 5 0. [] (iii) What does the value of this discriminant tell you about the roots of the equation x 4x 5 0? [1] (iv) Sketch the graph of y x 4x 5, showing the coordinates of the turning point and any points where the curve crosses the coordinate axes. [3] 3. (i) By factorising, solve the equation x x 6 0. [] (ii) Sketch the graph of y x x 6, showing the coordinates of any points where the graph cuts the coordinate axes. [3] 4. The quadratic equation x 5x k 0 has equal roots. (i) Find the value of k. [3] (ii) Solve the equation x 5x k 0. [] 5. (i) Write the expression x x 1 in the form (ii) Hence, or otherwise, solve the equation a( x p) q x x 1 0. [3]. [] 6. Sketch the graph of y 1 4x x, showing the coordinates of any points where the graph cuts the coordinate axes. [4] 7. Solve these equations, giving your answers in exact form. (i) (ii) x x 6 0 [4] 4 x 3x 10 0 [4] 8. The diagram shows a right-angled triangle. x Find the value of x, correct to 3 s.f. [4] x 9. Amy throws a ball so that when it is at its highest point, it passes through a hoop. 1 The path of the ball is modelled by the equation y h kx x, where y is the height of the ball above the ground and x is the horizontal distance from the point at which the ball was thrown. The centre of the hoop is at the point where x and y 5. Find the values of h and k, and find the value of x at which the ball hits the ground. [6] Total 50 marks 1 of 1 13/05/16 MEI integralmaths.org

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10 Notes and Examples AS Mathematics Trigonometry Section 3: Sine and cosine rules In this unit you learn about finding an unknown side or angle in any triangle. You will also learn a new formula for finding the area of a triangle. These notes contain subsections on: The sine rule The cosine rule Choosing which rule to use The area of a triangle The sine rule The sine rule: a b c sin A sin B sin C This form is easier to use when finding an unknown side. b A c The sine rule can also be written as: sin A sin B sin C a b c This form is easier to use when finding an unknown angle. C a B Note: When you use the sine rule to find a missing angle, θ, always check whether θ is a possible solution as well. Example 1 shows a straightforward application of the sine rule to find an unknown side. Example 1 Find the side BC in the triangle ABC. c B 30 x A 10 cm C 1 of 6 1/07/16 MEI

11 AS Maths Trigonometry 3 Notes and Examples Solution The triangle is isosceles so BAC is x b By the sine rule: sin A sin B x 10 So: sin 75 sin 30 10sin 75 x sin 30 so x = 19.3 cm (to 3 sig.fig.) Example shows a straightforward application of the sine rule to find an unknown angle. Example A, B and C are three points on a level plane. B is 6 km due west of A. C is 5 km from B and is on a bearing of 85 from A. Find ACB. Solution First draw a diagram: By the sine rule: So: a C 5 km Due west is a bearing of 70, so this angle must be B 6 km c sin A sin C a c sin15 sin C 5 6 6sin15 sin C 5 sin C C = 18.1 to 1 d.p. Check whether C is also a solution: = to 1 d.p. This also works so ACB is 18.1 to 1 d.p. or to 1 d.p b N A Your diagram doesn t need to be accurate just large enough to show all the information Don t round here! Store the number in your calculator. Angles A and C still add up to less than 180 of 6 1/07/16 MEI

12 AS Maths Trigonometry 3 Notes and Examples You can see examples similar to this using the Geogebra resource The sine rule finding an angle. This resource also shows geometrically what is happening when there is more than one possible solution. The cosine rule The cosine rule: A a b c bc cos A This form is easier to use when finding an unknown side. b c The cosine rule can also be written as: C b c a cos A bc This form is easier to use when finding an unknown angle. a B Example 3 shows an application of the cosine rule to find an unknown side. Example 3 Find the side YZ in the triangle XYZ. Z y 7 cm X 95 x z 6 cm Y Solution The cosine rule for this triangle is: So: x y z yz cos X x cos95 x x = 9.61 cm to 3 sig. fig. Example 4 shows a straightforward application of the cosine rule to find an unknown angle. 3 of 6 1/07/16 MEI

13 AS Maths Trigonometry 3 Notes and Examples Example 4 Find the angle θ in the triangle ABC. A 5 m b Solution The cosine rule for this triangle is: c 7 m θ B C 4 m a a b c cosc ab cosc 45 cos C 0. C = to 1 d.p. You can see examples similar to this using the Geogebra resource The cosine rule finding an angle. Choosing which rule to use Use the sine rule when: you know sides and 1 angle (not between the two sides) and want a nd angle (3 rd angle is now obvious!) you know angles and 1 side and want a nd side Use the cosine rule when: you know 3 sides and want any angle you know sides and the angle between them and want the 3 rd side Example 5 shows how to decide whether to use the sine or the cosine rule. Example 5 A ship sails from a port, P, 6 km due East to a lighthouse, L, 6 km away. The ship then sails 10 km on a bearing of 030 to an island, A. Find: (i) The distance AP (ii) The bearing of P from A 4 of 6 1/07/16 MEI

14 AS Maths Trigonometry 3 Notes and Examples Solution First draw a diagram: N N A P 6 km a l 10 L p 10 km 30 (i) You know sides and the angle between them so you need the cosine rule. The cosine rule for this triangle is: l a p ap cos l So the distance AP is 14 km. l cos10 l 196 l 14 km (ii) You can now use either the cosine rule or the sine rule to find the angle PAL. The sine rule for this triangle is: sin A sin L a l sin A sin10 So: sin10 sin A 14 sin A A = 1.8 So the bearing is = 31.8 to 1 d.p. Check whether = 158. is also a solution. It isn t because the angles in the triangle would total more than 180. The area of a triangle To find the area of any triangle you can use the rule: Area of triangle ABC 1 sin ab C So you need two sides and the angle between them. Example 6 shows how to use this formula. Example 6 Find the area of triangle ABC from Example 4. 5 of 6 1/07/16 MEI

15 Solution A AS Maths Trigonometry 3 Notes and Examples c 7 m 5 m b θ B C 4 m a In Example 4, angle C was found to be to 1 d.p. Using the formula Area = 1 absin C gives: 1 Area of triangle ABC = 4 5 sin m 6 of 6 1/07/16 MEI

16 Exercise level 1 AS Mathematics Trigonometry Section 3: The sine and cosine rules 1. Solve the triangle ABC in which A = 66, B = 4 and c = 1 cm.. Find two possible values of c in triangle ABC given that a = 16 cm, b = 10 cm, and B = Solve the triangle ABC in which a = 6 cm, b = 9 cm and C = Solve the triangle PQR in which p = 8 cm, q = 9 cm and r = 10 cm. 5. In triangle XYZ, X = 100, Y = 30 and XY = 10 cm. Calculate the area of the triangle. 6. The area of a triangle is 1 cm. Two of the sides are of lengths 6 cm and 7 cm. Calculate possible lengths for the third side. 7. A ship S is 6.8 km from a lighthouse on a bearing of 310. A second ship T is 8.4 km from the lighthouse on a bearing 075. Calculate ST and the bearing of T from S correct to the nearest degree. 8. Find all the lettered edges and angles in the figures in the following diagrams: (i) (ii) 10 o 6 a α 8 50 o 6 7 β b (iii) 0 d δ c 0 o 10 5 ε η (iv) φ 10 θ x 110 o 5 1 of 1 1/07/16 MEI

17 MEI AS Mathematics Trigonometry Section 3: The sine and cosine rules Exercise level 1. A golfer hits a ball B a distance of 170 m on a hole that measures 195 m from tee to hole. If his shot is directed 10 away from the direct line to the hole, find how far his ball is from the hole.. Calculate AB in the diagram below given that CD is 15 m, angle BCA = 40 and angle BDA = 0. A B C D 3. A tower stands on a slope inclined at 18 to the horizontal. From a point lower down the slope and 150 m from the base of the tower, the angle of elevation of the top of the tower is 7.5 (measured from the horizontal). Find the height of the tower. 4. A barge is moving at a constant speed along a straight canal. The angle of elevation of a bridge is 10. After 10 minutes the angle of elevation is 15. After how much longer does the barge reach the bridge? Give your answer to the nearest second. 5. Find all the lettered edges and angles in the figures in the following diagrams: (i) a b 15 o 40 o 10 x h (ii) α 6 β θ β is acute 1 x φ 0 o 30 o c 1 of 1 integralmaths.org 01/11/17 MEI

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19 Notes and Examples AS Mathematics Problem solving Section : Notation and proof These notes contain subsections on Mathematical language The converse of a theorem Proof Mathematical language In this section you are introduced to the symbols, and. You have probably seen these used before they are used in most mathematics textbooks at this level. You will probably find using the symbols, and fairly straightforward. The use of the words necessary and sufficient may be a little harder to understand. Here are a few examples which may help you. You may need to read them through more than once! 1. A number ends in 5 the number is divisible by 5. A number ends in 5 is a sufficient condition for the number is divisible by 5, since there are no numbers which end in 5 which are not divisible by 5. However, it is not a necessary condition, there are numbers which do not end in 5 which are divisible by 5 (numbers which end in zero). You can express this the other way round: A number is divisible by 5 the number ends in 5. A number is divisible by 5 is a necessary condition for the number ends in 5, since all numbers which end in 5 are divisible by 5. However, it is not a sufficient condition, as not all numbers divisible by 5 end in 5.. A number is even the number is divisible by 4. A number is even is a necessary condition for the number is divisible by 4, since all numbers which are divisible by 4 are even. However, it is not a sufficient condition, as not all even numbers are divisible by 4. Again, you can write this the other way round: A number is divisible by 4 the number is even. A number is divisible by 4 is a sufficient condition for the number is even, since there are no numbers which are divisible by 4 which are not even. 1 of 4 17/05/16 MEI

20 AS Maths Problem solving Notes and examples However, it is not a necessary condition, as there are even numbers which are not divisible by A number is divisible by 10 the number ends in a zero. A number is divisible by 10 is a necessary and sufficient condition for the number ends in zero. All numbers which are divisible by 10 end in zero, and all numbers which end in zero are divisible by 10. Another way of expressing this is the statement A number is divisible by 10 if and only if the number ends in zero. In the table below, all the statements shown in the same column are equivalent to each other. A is a necessary condition for B If B is true, then A must also be true A B A is implied by B or A follows from B A is a sufficient condition for B If A is true, then B must also be true A B A implies B A is a necessary and sufficient condition for B A is true if and only if B is true. A B A implies and is implied by B You can often use the symbols, and when you are writing out a solution to a mathematical problem. For example, if you want to multiply out the equation y (x 3)( x 1), you might write: y (x 3)( x 1) y x x x 3 3 y x x 3 In fact, you could use instead of. y (x 3)( x 1) y x x x 3 3 y x x 3 People often use when they could use, if they are only interested in the logical steps in one direction. of 4 17/05/16 MEI

21 AS Maths Problem solving Notes and examples The converse of a theorem It should be clear from the examples above that just because a theorem, or a statement, is true, does not necessarily mean that the converse is also true. If a theorem can be written using, then both the theorem and its converse are true. For example: A triangle has equal sides the triangle has equal angles. Both this statement and its converse are true. The statement An equation is linear the equation has exactly one real root is true, since all linear equations have exactly one real root. However, the converse of this statement would be An equation has exactly one real root the equation is linear which is not true, since there are many examples of quadratics, cubics and indeed polynomial equations of any order which have exactly one real root. This statement cannot be written using. Proof Proof is a very important aspect of mathematics and you are expected to have an idea about what proof involves. The most important thing to realise is that checking lots of cases does not prove that the result is true. As an example, think of three consecutive even numbers (such as 4, 6 and 8 or, 4 and 6) and add them up. You should find that this sum is divisible by 6. Suppose you want to prove that the sum of three consecutive even integers is always divisible by 6. You could test quite a lot of sets of numbers yourself, or you could program a computer to test a very large number of sets of numbers. The computer could keep checking numbers up to astronomically large numbers, but you would still not have checked every single number, and you never can, since there are an infinite number of sets of three consecutive integers! At this stage you could feel sure that the conjecture is in fact true, but to prove it you need to show that it is true for all possible sets of numbers. Fortunately, there is a method for this. 3 of 4 17/05/16 MEI

22 AS Maths Problem solving Notes and examples Example 1 Prove that the sum of any three consecutive even integers is divisible by 6. Solution Any even number is twice a whole number so let the first number be n, where n is an integer. Then the second number is n +, and the third number is n + 4. The sum of the three numbers is n + n + + n + 4 = 6n + 6 = 6(n + 1) 6(n + 1) is divisible by 6 for all values of n. This is an example of direct proof, or proof by deduction. In this case the proof consists of a set of logical steps. Proof by exhaustion, where there are a limited number of possibilities which can all be tested, is another method of proof. Here is an example Example Prove that even number from between 4 and 50 is the sum of two prime numbers. (Note - is conjectured but not proved that every even number is the sum of two primes this is known as Goldbach s conjecture.) Solution This is shown directly for every case here: 4 = + 6 = = = = = = = = = = = = = = = = = = = = = = = To disprove a conjecture, all you need is to find a single instance where the conjecture is not true. This is called a counter example. Here is a conjecture which is not true: For any positive integer n, the sum of any n consecutive integers is a multiple of n. Here is a counter example which disproves it. With n = 4, = 10 is the sum of 4 consecutive integers but this is not a multiple of 4. 4 of 4 17/05/16 MEI

23 Exercise level 1 AS Mathematics Problem solving Section : Notation and proof The following table contains pairs of statements, A and B. In each case complete the central column with one of the three symbols,,, or with none. Also in each case, add the letters N (necessary), S (sufficient), or N & S (necessary and sufficient), or neither, to indicate the relationship between the statements. Q. no. A nec. or suff. for B? 1 Statement A,,, none? Statement B x 9 x < 6 B nec. or suff. for A? This month has 9 days It is February in a leap year 3 x 1 or x 1 ( x1)( x1) 0 4 sin x 1 x 90 5 A polygon is a square A polygon has 4 sides 6 ( x1)( y1) 0 x 1 and y 1 7 Three straight lines in d meet in exactly points Exactly of 3 straight lines in d are parallel 8 x 9 x = 3 9 n is a multiple of 5 n is a multiple of 5 10 k is divisible by 3 k + 1 is even 11 The discriminant of a quadratic equation is non-negative A quadratic equation has unequal roots 1 a b 1 1 a b 1 of 1 17/05/16 MEI

24 Exercise level AS Mathematics Problem solving Section : Notation and proof The idea of proof and the correct use of the language of mathematics is fundamental to success in maths. As you will have seen in your study, there are many ways of proving or disproving mathematical statements: Statements can be proved by various different methods including exhaustion (trying everything until you have used up all the possibilities) and logical deduction (often using algebra, based on previous knowledge) Statements are often disproved by a simple counter-example (often very difficult to find!) As you work through these exercises, the important thing is to concentrate on the logic and the language you do not need to learn any of these proofs. 1. (deduction) Prove that if n is an even number then n² + n is an even number. (deduction) Prove that (i) the sum of five consecutive integers is a multiple of 5. (ii) the mean of five consecutive integers is equal to the third number. 3. (counter example) The following statements are all false. Find a counter example that disproves them in each case (i) the difference between any two square numbers is odd (ii) all prime numbers are odd (iii) for any positive integer n, the number n( n 1) ( n )... 1 has n 1 distinct factors 4. (exhaustion) Prove that if the name of a month has 5 or more characters, then a 4-letter word can be formed using those characters. 1 of 1 16/05/16 MEI

25 Answers Quadratic Equations 1(i) 1ii) impossible 1(iii)

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27 Trigonometry Exercise Level 1 1. a=11.53cm b=8.44cm. A= or C= or c=19.9cm or 7.9cm 3 C=11.4cm, A= B= P= Q= R= Z=50 0 x=1.86cm area of the triangle = 3.1cm 6. The possible length of the side are 4.01cm or 1.41cm 7. ST=13.5km bearing of T from S = (i) , 13.8 (ii) b=5.57, B=74.3 (iii) c=7.8, d=11.1,, Exercise Level m m m minutes 15 seconds 5. (i)x=4.69, h=3.94 (ii)x=15.65, c=47. Proofs and Mathematical Notation Most of the questions here can be done in several ways. Try to come up with your own methods of doing the proofs. We will discuss the solutions in September.

28 Mini Personalised Learning checklist Topic Factorising quadratics Completing the square Solving quadratic Equations Quadratic equations ( word problems) Solving simultaneous Equations Simultaneous equations (word problems) Sine and cosine rule Sine and cosine rule (applications, e.g. bearing) Mathematical notations Using different techniques of proofs RAG